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# JEE Main 2019 Question Paper with Solutions (11th January - Morning)

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JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 1

### A body is projected at t = 0 with a velocity 10 ms-1 at an angle of 60° with the horizontal. The radius of curvature of its trajectory at t = 1s is R. Neglecting air resistance and taking acceleration due to gravity g = 10 ms-2, the value of R is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 1

vx = 10cos60° = 5 m/s
vy = 10cos30° = 5 √3 m/s
velocity after t = 1 sec.
vx = 5 m/s

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 2

### A particle is moving along a circular path with a constant speed of 10 ms-1. What is the magnitude of the change in velocity of the particle, when it moves through an angle of 60° around the centre of the circle?

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 2

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 3

### A hydrogen atom, initially in the ground state is excited by absorbing a photon of wavelength 980Å. The radius of the atom in the excited state, in terms of Bohr radius a0, will be : (hc = 12500 eV – Å)

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 3

Energy of photon = 12500/980 = 12.75eV
∴ Electron will excite to n= 4
Since 'R' ∝  n2
∴ Radius of atom will be 16a0

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 4

A liquid of density ρ is coming out of a hose pipe of radius a with horizontal speed v and hits a mesh. 50% of the liquid passes through the mesh unaffected. 25% looses all of its momentum and 25% comes back with the same speed. The resultant pressure on the mesh will be :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 4

Momentum per second carried by liquid per second is ρav2
net force due to reflected liquid =
net force due to stopped liquid =

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 5

An electromagnetic wave of intensity 50 Wm-2 enters in a medium of refractive index 'n' without any loss. The ratio of the magnitudes of electric fields, and the ratio of the magnitudes of magnetic fields of the wave before and after entering into the medium are respectively, given by :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 5

similarly

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 6

An amplitude modulated signal is given by V(t) = 10[1 + 0.3cos(2.2 x 104)]sin(5.5 x 105t). Here t is in seconds. The sideband frequencies (in kHz) are, [Given π = 22/7]

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 6

Side band frequency are

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 7

The force of interaction between two atoms is given by where x is the distance, k is the Boltzmann constant and T is temperature and α and β are two constants. The dimension of β is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 7

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 8

The charges Q + q and +q are placed at the vertices of a right-angle isosceles triangle as shown below. The net electrostatic energy of the configuration is zero, it the value of Q is:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 8

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 9

In the circuit shown, the switch S1 is closed at time t = 0 and the switch S2 is kept open. At some later time (t0), the switch S1 is opened and S2 is closed. The behaviour of the current I as a function of time 't' is given by :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 9

From time t = 0 to t = t0, growth of current takes place and after that decay of current takes place.

most appropriate is (2)

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 10

Equation of travelling wave on a stretched string of linear density 5 g/m is y = 0.03 sin(450 t – 9x) where distance and time are measured is SI units. The tension in the string is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 10

y = 0.03 sin(450 t – 9x)

⇒ T = 2500 × 5 × 10–3
= 12.5 N

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 11

An equilateral triangle ABC is cut from a thin solid sheet of wood. (see figure) D, E and F are the mid-points of its sides as shown and G is the centre of the triangle. The moment of inertia of the triangle about an axis passing through G and perpendicular to the plane of the triangle is I0. It the smaller triangle DEF is removed from ABC, the moment of inertia of the remaining figure about the same axis is I. Then:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 11

Suppose M is mass and a is side of larger
triangle, then M/4 and a/2 will be mass and side length of smaller triangle.

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 12

There are two long co-axial solenoids of same length l. the inner and outer coils have radii r1 and r2 and number of turns per unit length n1 and n2 respectively. The ratio of mutual inductance to the self-inductance of the inner-coil is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 12

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 13

A rigid diatomic ideal gas undergoes an adiabatic process at room temperature,. The relation between temperature and volume of this process is TVx = constant, then x is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 13

For adiabatic process : TVgγ-1 = constant
For diatomic process :

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 14

The gas mixture constists of 3 moles of oxygen and 5 moles of argon at temperature T. Considering only translational and rotational modes, the total inernal energy of the system is:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 14

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 15

In a Young's double slit experiment, the path different, at a certain point on the screen, between two interfering waves is 1/8th of wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is close to :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 15

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 16

If the deBronglie wavelength of an electron is equal to 10–3 times the wavelength of a photon of frequency 6 × 1014 Hz, then the speed of electron is equal to :
(Speed of light = 3 × 108 m/s
Planck's constant = 6.63 × 10–34 J.s
Mass of electron = 9.1 × 10–31 kg)

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 16

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 17

A slab is subjected to two forces of same magnitude F as shown in the figure. Force is in XY-plane while force F1 acts along z-axisat the point . The moment of these forces about point O will be :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 17

Torque for F1 force

Torque for F2 force

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 18

A satellite is revolving in a circular orbit at a height h from the earth surface, such that h << R where R is the radius of the earth.
Assuming that the effect of earth's atmosphere can be neglected the minimum increase in the speed requried so that the satellite could escape from the gravitational field of earth is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 18

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 19

In an experiment electrons are accelerated, from rest, by applying a voltage of 500 V. Calculate the radius of the path if a magnetic field 100 mT is then applied.
[Charge of the electron = 1.6 × 10–19 C Mass of the electron = 9.1 × 10–31 kg]

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 19

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 20

A particle undergoing simple harmonic motion has time dependent displacement given by The ratio of kinetic to potential energy of this particle at t = 210 s will be :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 20

Hence ratio is 3 (most appropriate)

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 21

Ice at –20°C is added tp 50 g of water at 40°C. When the temperature of the mixture reaches 0°C, it is found that 20 g of ice is still unmelted. The amount of ice added to the water was close to
(Specific heat of water = 4.2 J/g/°C)
Specific heat of Ice = 2.1 J/g/°C
Heat of fusion of water at 0°C = 334 J/g)

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 21

Let amount of ice is m gm.
According to principal of calorimeter
heat taken by ice = heat given by water
∴ 20 × 2.1 × m + (m - 20) × 334
= 50 × 4.2 × 40
376 m = 8400 + 6680
m = 40.1 g

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 22

In the figure shown below, the charge on the left plate of the 10 μF capacitor is -30 μC. The charge on the right plate of the 6 μF capacitor is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 22

6µF & 4µF are in parallel & total charge on this combination is 30 µC
∴ Charge on 6µF capacitor =
= 18 µC
Since charge is asked on right plate therefore is +18µC
Correct answer is (4)

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 23

In the given circuit the current through Zener Diode is close to :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 23

Since voltage across zener diode must be less than 10V therefore it will not work in breakdown region, & its resistance will be infinite & current through it = 0
∴ correct answer is (4)

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 24

The variation of refractive index of a crown glass thin prism with wavelength of the incident light is shown. Which of the following graphs is the correct one, if Dm is the angle of minimum deviation?

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 24

Since Dm = (µ – 1)A
& on increasing the wavelength, µ decreases & hence Dm decreases. Therefore correct answer is (2)

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 25

The resistance of the meter bridge AB in the given figure is 4Ω. With a cell of emf ε = 0.5 V and rheostat resistance Rh = 2Ω the null point is obtained at some point J. When the cell is replaced by another one of emf ε = ε2 the same null point J is found for Rh = 6 Ω. The emf ε2 is;

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 25

Potential gradient with Rh = 2Ω

Let null point be at ℓ cm

Now with Rh = 6Ω new potential gradient is  and at null point
dividing equation (1) by (2) we get

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 26

The given graph shows variation (with distance r from centre) of :

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 27

Two equal resistance when connected in series to a battery, consume electric power of 60 W. If these resistances are now connected in parallel combination to the same battery, the electric power consumed will be :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 27

In series condition, equivalent resistance is 2R thus power consumed is
In parallel condition, equivalent resistance is R/ 2 thus new power is

o r P' = 4P = 240W

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 28

An object is at a distance of 20 m from a convex lens of focal length 0.3 m. The lens forms an image of the object. If the object moves away from the lens at a speed of 5 m/s, the speed and direction of the image will be :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 28

From lens equation

velocity of image wrt. to lens is given by vI/L = m2vO/L
direction of velocity of image is same as that of object
vO/L = 5 m/s

= 1.16 × 10–3 m/s towards the lens

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 29

A body of mass 1 kg falls freely from a height of 100 m on a platform of mass 3 kg which is mounted on a spring having spring constant k = 1.25 x 106 N/m. The body sticks to the platform and the spring's maximum compression is found to be x. Given that g = 10 ms-2, the value of x will be close to :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 29

Velocity of 1 kg block just before it collides with 3kg block =
Applying momentum conversation just before and just after collision.

initial compression of spring 1.25 × 106 x0 = 30 ⇒  x0 ≈  0
applying work energy theorem,Wg + Wsp = ΔKE

solving x ≈ 4 cm

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 30

In a Wheatstone bridge (see fig.), Resistances P and Q are approximately equal. When R = 400 Ω, the bridge is equal. When R = 400 Ω, the bridge is balanced. On inter-changing P and Q, the value of R, for balance, is 405 Ω. The value of X is close to :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 30

After interchanging P and Q

From (i) and (ii)

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 31

For the cell Zn(s) | Zn2+(aq) || Mx+ (aq) | M(s), different half cells and their standard electrode potentials are given below :

If , which cathode will give a maximum value of E0cell per electron transferred?

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 31

We have,

E0cell​= E0cathode−E0​anode0​

For a high value of E0​cell the value of SRP of cathode should be high.

Here the highest value is for Au3+/Au

E0cell= 1.4−(−0.76)=2.16V

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 32

The correct match between items-I and II is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 32

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 33

If a reaction follows the Arrhenius equation, the plot lnk vs 1/(RT) gives straight line with agradient (–y) unit. The energy required to activate the reactant is :

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 34

The concentration of dissolved oxygen (DO) in cold water can go upto :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 34

In cold water, dissolved oxygen (DO) can reach a concentration upto 10 ppm

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 35

The major product of the following reaction is:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 35

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 36

Th correct statements among (a) to (d) regarding H2 as a fuel are :
(a) It produces less pollutant than petrol
(b) A cylinder of compressed dihydrogen weighs ~ 30times more than a petrol tank producing the same amount of energy
(c) Dihydrogen is stored in tanks of metal alloys like NaNi5
(d) On combustion, values of energy released per gram of liquid dihydrogen and LPG are 50 and 142 kJ, respectively

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 37

The major poduct of the following reaction is:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 37

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 38

The element that usually does not show variable oxidation states is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 38

Usally Sc(Scandium) does not show variable oxidation states.
Most common oxidation states of :
(i) Sc : +3
(ii) V : +2, +3, +4, +5
(iii) Ti : +2, +3, +4
(iv) Cu : +1, +2

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 39

An organic compound is estimated through Dumus method and was found to evolve 6 moles of CO2. 4 moles of H2O and 1 mole of nitrogen gas. The formula of the compound is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 39

Hence, C6H8N2

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 40

The major product of the following reaction is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 40

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 41

Among the following compound which one is found in RNA?

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 41

For the given structure 'uracil' is found in RNA

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 42

Which compound(s) out of the following is/are not aromatic?

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 42

out of the given options only  is aromatic.
Hence (B), (C) and (D) are not aromatic

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 43

The correct match between Item (I) and Item (II) is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 43

(A) Norethindrone – Antifertility
(B) Ofloaxacin – Anti-Biotic
(C) Equanil – Hypertension (traiquilizer)

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 44

Heat treatment of muscular pain involves radiation of wavelength of about 900 nm. Which spectral line of H-atom is suitable for this purpose?
[RH = 1 × 105 cm–1, h = 6.6 × 10–34 Js, c = 3 × 108 ms–1]

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 45

Consider the reaction,
N2(g) + 3H2(g) ⇔ 2NH3(g)
The equilibrium constant of the above reaction is KP. If pure ammonia is left to dissociate, the partial pressure of ammonia at equilibrium is given by
(Assume that PNH3 <<Ptotal at equilibrium)

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 46

Match the ores(Column A) with the metals (column B) :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 46

Siderite : FeCO3
Kaolinite : Al2(OH)4Si2O5
Malachite : Cu(OH)2.CuCO3
Calamine : ZnCO3

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 47

The correct order of the atomic radii of C, Cs, Al and S is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 47

Atomic radii order : C < S < Al < Cs
Atomic radius of C : 170 pm
Atomic radius of S : 180 pm
Atomic radius of Al : 184 pm
Atomic radius of Cs : 300 pm

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 48

Match the metals (Column I) with the coordination compound(s) / enzyme(s) (Column II)

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 48

(i) Wilkinson catalyst : RhCl(PPh3)3
(ii) Chlorophyll : C55H72O5N4Mg
(iii) Vitamin B12(also known as cyanocobalamin) contain cobalt.
(iv) Carbonic anhydrase  contains a zinc ion.

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 49

A 10 mg effervescent tablet contianing sodium bicarbonate and oxalic acid releases 0.25 ml of CO2 at T = 298.15 K and p = 1 bar. If molar volume of CO2 is 25.0 L under such condition, what is the percentage of sodium bicarbonate in each tablet ?
[Molar mass of NaHCO3 = 84 g mol-1]

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 49

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 50

The major product of the following reaction is:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 50

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 51

Two blocks of the same metal having same mass and at temperature T1 and T2, respectively. are brought in contact with each other and allowed to attain thermal equilibrium at constant pressure. The change in entropy, ΔS, for this process is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 51

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 52

The chloride that CANNOT get hydrolysed is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 52

CCl4 cannot get hydrolyzed due to the absence of vacant orbital at carbon atom.

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 53

For the ch emical reaction X ⇔ Y, the standard reaction Gibbs energy depends on temperature T (in K) as :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 53

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 54

The freezing point of a diluted milk sample is found to be –0.2°C, while it should have been –0.5°C for pure milk. How much water has been added to pure milk to make the diluted sample?

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 54

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 55

A solid having density of 9 × 103 kg m–3 forms face centred cubic crystals of edge length 200√2 pm. What is the molar mass of the solid ?
(Avogadro constant ≌ 6 x 1023 mol-1, π ≌ 3)

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 56

The polymer obtained from the following reactions is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 56

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 57

An example of solid sol is :

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 58

Peroxyacetyl nitrate (PAN), an eye irritant is produced by :​

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 58

Photochemical smog produce chemicals such as formaldehyde, acrolein and peroxyacetyl nitrate (PAN).

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 59

NaH is an example of :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 59

NaH is an example of ionic hydride which is also known as saline hydride.

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 60

The amphoteric hydroxide is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 60

Be(OH)2 is amphoteric in nature while rest all alkaline earth metal hydroxide are basic in nature.

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 61

Let  It AAT = I3, then |p| is

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 61

A is orthogonal matrix

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 62

The area (in sq. units) of the region bounded by the curve x2 = 4y and the straight line x = 4y – 2 :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 62

x = 4y – 2 & x2 = 4y
⇒ x2 = x + 2 ⇒ x2 – x – 2 = 0
x = 2, – 1

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 63

The outcome of each of 30 items was observed; 10 items gave an outcome 1/2 -d each, 10 items gave outcome 1/2 each and the remaining 10 items gave outcome 1/2 + d each. If the variance of this outcome data is 4/3 then |d| equals :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 63

Variance is independent of origin. So we shift the given data by 1/2.

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 64

The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of its terms is 27/19. Then the common ratio of this series is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 64

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 65

Let and  be coplanar vectors. Then the non-zero vector  is:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 65

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 66

Let , where x and y are real numbers, then y – x equals :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 66

Hence, y – x = 198 – 107 = 91

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 67

Let  and g(x) = |f(x)| + f (|x|). Then, in the interval (–2, 2), g is :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 67

and f(|x|) = x2 - 1, x ∈ [-2, 2]

It is not differentiable at x = 1

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 68

Let f : R → R be defined by x ∈ R. Then the range of f is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 68

f(0) = 0 & f(x) is odd.
Further, if x > 0 then

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 69

The sum of the real values of x for which the middle term in the binomial expansion of equals 5670 is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 69

⇒ 70x8 = 5670

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 70

The value of r for which 20Cr 20C0 + 20Cr–1 20C1 + 20Cr–2 20C2 + .... 20C0 20Cr  is maximum, is

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 70

Given sum = coefficient of xr in the expansion of (1 + x)20(1 + x)20,
which is equal to 40Cr
It is maximum when r = 20

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 71

Let a1, a2, ....., a10 be a G.P. If a3/a1 = 25, then a9/a5 equals.

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 71

a1, a2, ....., a10 are in G.P.,
Let the common ratio be r

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 72

If  for a suitable chosen integer m and a function A(x), where C is a constant of integration then (A(x))m equals :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 72

Case-II x ≤ 0

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 73

In a triangle, the sum of lengths of two sides is x and the product of the lengths of the same two sides is y. If x2 - c2 = y, where c is the length of the third side of the triangle, then the circumradius of the triangle is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 73

Given a + b = x  and ab = y
If x2 – c2 = y ⇒ (a + b)2 – c2 = ab
⇒ a2 + b2 – c2 = –ab

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 74

The value of the integral (where [x] denotes the greatest integer less than 20Cr or equal to x) is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 74

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 75

If the system of linear equations
2x + 2y + 3z = a
3x - y + 5z = b
x - 3y + 2z = c
where a, b, c are non-zero real numbers, has more then one solution, then :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 75

P1 : 2x + 2y + 3z = a
P2 : 3x – y + 5z = b
P3 : x – 3y + 2z = c
We find
P1 + P3 = P2 ⇒ a + c = b

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 76

A square is inscribed inthe circle x2 + y2 – 6x + 8y – 103 = 0 with its sides parallel to the corrdinate axes. Then the distance of the vertex of this square which is nearest to the origin is :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 76

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 77

Let fk(x) = 1/k (sinx + cosk x) for k = 1,2,3,....... Then for all x ∈ R, the value of f4(x) – f6(x) is equal to :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 77

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 78

Let [x] denote the greatest integer less than or equal to x. Then :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 78

R.H.L. ≠ L.H.L.

*Multiple options can be correct
JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 79

The direction ratios of normal to the plane through the points (0, –1, 0) and (0, 0, 1) and making an anlge π/4 with the plane y–z+5=0 are:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 79

Let the equation of plane be a(x – 0) + b(y + 1) + c(z – 0) = 0 It passes through (0,0,1) then
b + c = 0 ...(1)

⇒ a2 = –2bc and b = –c
we get a2 = 2c2

⇒ direction ratio (a, b, c) = (√2, -1, 1) or  (√2, 1, -1)

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 80

If xloge(loge x) – x2 + y2 = 4(y > 0), then dy/dx at x = e is equal to :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 80

Differentiating with respect to x,

at x = e we get

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 81

The straight line x + 2y = 1 meets the coordinate axes at A and B. A circle is drawn through A, B and the origin. Then the sum of perpendicular distances from A and B on the tangent to the circle at the origin is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 81

Equation of circle

Equation of tangent of origin is 2x + y = 0

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 82

If q is false and p ∧ q⇔r is true, then which one of the following statements is a tautology?

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 82

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 83

If y(x) is the solution of the differential equation where, then :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 83

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 84

The maximum value of the function f(x) = 3x3 – 18x2 + 27x – 40 on the set S = {x ∈ R : x2 + 30 ≤ 11x} is:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 84

S = {x ∈ R, x2 + 30 – 11x ≤ 0}
= {x ∈ R, 5 ≤ x ≤ 6}
Now f(x) = 3x3 – 18x2 + 27x – 40
⇒ f'(x) = 9(x – 1)(x – 3),
which is positive in [5, 6]
⇒ f(x) increasing in [5, 6]
Hence maximum value = f(6) = 122

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 85

If one real root of the quadratic equation 81x2 + kx + 256 = 0 is cube of the other root, then a value of k is

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 85

81x2 + kx + 256 = 0 ; x = α, α3

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 86

Two circles with equal radii are intersecting at the points (0, 1) and (0, –1). The tangent at the point (0, 1) to one of the circles passes through the centre of the other circle. Then the distance between the centres of these circles is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 86

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 87

Equation of a common tangent to the parabola y2 = 4x and the hyperbole xy = 2 is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 87

Let the equation of tangent to parabola
y2 = 4x be y = mx + 1/m
It is also a tangent to hyperbola xy = 2

So tangent is 2y + x + 4 = 0

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 88

The plane containing the line  and also containing its projection on the plane 2x + 3y – z = 5, contains which one of the following points ?

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 88

The normal vector of required plane

So, direction ratio of normal is (–1, 1, 1) So required plane is
–(x – 3) + (y + 2) + (z – 1) = 0
⇒ –x + y + z + 4 = 0
Which is satisfied by (2, 0, –2)

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 89

If tangents are drawn to the ellipse x2 + 2y2 = 2 at all points on the ellipse other than its four vertices then the mid points of the tangents intercepted betwen the coordinate axes lie on the curve :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 89

Equation of general tangent on ellipse

Let the midpoint be (h, k)

JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 90

Two integers are selected at random from the set {1, 2,...., 11}. Given that the sum of selected numbers is even, the conditional probability that both the numbers are even is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Morning) - Question 90

Since sum of two numbers is even so either both are odd or both are even. Hence number of elements in reduced samples space
= 5C2 + 6C2
so required probability =

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